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Solutions to Problems 55 Propagation: H H (1) H-C-H H H :0: H :0: (2) S C-H + CH₃ :0: H :0: We will be stuck unless we can find a way for to give rise to another propagation step to con- tinue the chain process. Two options are possible, and with the information you have been given, either one is a reasonable proposal. (a) Unimolecular decomposition of ; that is, :0: S + Cl:, followed by a new propagation step 1, or (b) An alternative propagation step that uses in place of the CI in step 1: :0: H H S Cl: :0: H H Either is a qualitatively sensible mechanistic possibility. 36. Recall (text Section 3-4) that = 4 kcal mol = 4000 cal for the reaction between and CH₄. Therefore, + = A e-4000/(1.986)(298), and + = A So, + + = 15000/(1.986)(298) = = 9.7 X 10¹⁰. That's a pretty big rate ratio. 37. As we saw in the case of the reaction of methane with a mixture of Br₂ and Cl₂, the only kinetically viable first propagation steps in this case are reactions of atoms with propane. Reactions of Br atoms are far too slow to compete. It is this first step that determines the ratio of primary to secondary alkyl radicals that form. Therefore, the selectivity observed is that of atoms. The two radicals both proceed to react rapidly with either molecular halogen, Cl₂ or Br₂. Because the ratio of radicals present was determined in the prior step, the ratios of chloropropane isomers and of bromopropane isomers obtained are essentially the same, and reflect the selectivity of chlorination. 38. Calculate using the same method that was introduced in Section 3-6. There are three groups of hydrogens with different reactivities: two on C1, two on and three on C3. From the relative yields, it appears that those on C3 are lowest in reactivity. It is sensible therefore to calculate how much more reactive the hydrogens on C2 and are compared to those on C3. First, let's do C2 C3: Relative reactivity 8.5% of a hydrogen on C2 = = 8.5 Relative reactivity 1.5% C3 of a hydrogen on C3 chlorination, )/( C3